A farmer has 54 m of fencing with which to build two rectangular animal pens with a common side. If the area of the pens if to be maximized, what are their dimensions?

let the width of the whole thing be x m

and let the length be y m

your sketch should show 3 widths and 2 lengths
where 3x + 2y = 54
y = (54-3x)/2 = 27 - (3/2)x

area = xy
x(27 - (3/2)x)
= 27x - (3/2)x^2

d(area)/dx = 27 - 3x
= 0 for a max/min of area
3x = 27
x = 9
y = (54-27)/2 = 13.5

the whole field is 13.5 m by 9 m, with a 9 m divider as shown in your diagram.

To maximize the area of the two rectangular animal pens, we can set up the problem by dividing the available fencing into three parts: two equal lengths for the length of each pen, and one shorter length for the shared side.

Let's call the length of each rectangular pen "x" and the common side "y".

Given that the farmer has 54 meters of fencing:
2x + y = 54 (equation 1)

The total area of the two rectangular pens is the sum of the areas of each pen: A = x*y + x*y

To find the dimensions that maximize the area, we need to maximize the formula A = 2xy.

Now let's solve the problem using the steps below:

Step 1: Solve equation 1 for y:
y = 54 - 2x

Step 2: Substitute the value of y in terms of x into the formula for the area:
A = 2x(54 - 2x)

Step 3: Expand and simplify the equation:
A = 108x - 4x^2

Step 4: To find the maximum area, we differentiate A with respect to x and set it equal to zero:
dA/dx = 108 - 8x = 0

Step 5: Solve the equation for x:
108 - 8x = 0
8x = 108
x = 108 / 8
x = 13.5

Step 6: Substitute the value of x back into equation 1 to find y:
2(13.5) + y = 54
27 + y = 54
y = 54 - 27
y = 27

Therefore, the dimensions of the two rectangular animal pens that maximize the area are 13.5 meters by 27 meters.

To find the dimensions of the rectangular animal pens with the maximum area, we can use the derivative of the area function with respect to one of the dimensions.

Let's denote the length of the common side between the two pens as 'x', and the dimensions of the individual pens as 'l' and 'w'.

We are given that the farmer has 54 meters of fencing, which we can use to form the perimeter of the pens. The perimeter consists of the lengths of all four sides, so we can write the equation:

2l + x + w + x = 54

Simplifying the equation, we have:

2l + 2x + w = 54

Next, we need to express the area of the pens in terms of 'l' and 'w'. The area of a rectangle is given by:

A = l * w

Since we have two pens, the total area will be twice the area of a single pen:

A_total = 2 * A

Substituting 'l' and 'w' in terms of 'x' from the perimeter equation, we have:

A_total = 2 * (l * w)
= 2 * (x * (54 - 2x - w))

To find the dimensions that maximize the area, we can differentiate the area function with respect to 'x':

dA_total/dx = 2 * (54 - 4x - w)

Setting the derivative equal to zero to find the critical points:

2 * (54 - 4x - w) = 0

54 - 4x - w = 0
w = 54 - 4x

Substituting this value of 'w' back into the perimeter equation:

2l + 2x + (54 - 4x) = 54
2l - 2x = 0
l = x

Therefore, the dimensions of the pens that maximize the area are:

length: l = x
width: w = 54 - 4x

To find the specific dimensions, we need to solve the system of equations. Substituting 'l = x' into the perimeter equation:

2(x) + x + (54 - 4x) + x = 54
4x = 54
x = 13.5

Now we can find the dimensions of the pens:

length: l = x = 13.5 meters
width: w = 54 - 4x = 54 - 4(13.5) = 54 - 54 = 0 meters

Therefore, the dimensions of the pens that maximize the area are:

length: 13.5 meters
width: 0 meters

However, the width of 0 meters is not physically feasible, so we can conclude that one of the pens should have dimensions 13.5 meters by 0 meters, while the other pen should have dimensions 0 meters by 13.5 meters.